Question

9 064g Imol = 1.88x10th mol 340.19 reactant 1 (E-stilbenel o.osog 1 mol = 2.77 x 16 mol 1180.2g B reactant 2 (pyridinium trib

How do you find the limiting reactant? I know the formula, but I cant remember how to round the moles to a whole number.
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Answer #1

A+ B c . Moles of a formed = 1.888104 mel. Moles of B reacted = 1.888104 mol. Moles of A reached = 1:88x154 mol. Moles of A u

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